Ask question

Ask question

All categories

3 years ago

6

Please answer the Scientific Notation Question! And if you can, please also generously explain. Your help is greatly appreciated

!

1 answer:

Ugo [173]3 years ago

8 0

Convert the fraction to a decimal by dividing the numerator by the denominator.

1.05*10^-10

so a is the answer

You might be interested in

Item 5<br> Write 77.5% as a fraction or mixed number in simplest form.

jeyben [28]

<span>77.5% can also be written as 77.5/100, but if you're the mathematical notation police, people don't like decimals on a fraction, sooo, an equivalent form is: </span>

<span>
= (77.5)/(100) * (2/2)
= 155/200
= 31/40
= 0 31/40 </span>

5 0

2 years ago

In 20 minutes a heart can beat 700 times. At this rate in how many minutes will a heart beat 140 times at what rate can a heart

murzikaleks [220]

4900 is the answer :)

5 0

3 years ago

Find the sum of (4x^3+2x) + (8x^3+4)

Savatey [412]

The answer would be 12x^3+ 2x + 4

5 0

3 years ago

Find two power series solutions of the given differential equation about the ordinary point x = 0. compare the series solutions

monitta

I don't know what method is referred to in "section 4.3", but I'll suppose it's reduction of order and use that to find the exact solution. Take $z=y'$ , so that $z'=y''$ and we're left with the ODE linear in $z$ :

$y''-y'=0\implies z'-z=0\implies z=C_1e^x\implies y=C_1e^x+C_2$

Now suppose $y$ has a power series expansion

$y=\displaystyle\sum_{n\ge0}a_nx^n$
$\implies y'=\displaystyle\sum_{n\ge1}na_nx^{n-1}$
$\implies y''=\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}$

Then the ODE can be written as

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-\sum_{n\ge1}na_nx^{n-1}=0$

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-\sum_{n\ge2}(n-1)a_{n-1}x^{n-2}=0$

$\displaystyle\sum_{n\ge2}\bigg[n(n-1)a_n-(n-1)a_{n-1}\bigg]x^{n-2}=0$

All the coefficients of the series vanish, and setting $x=0$ in the power series forms for $y$ and $y'$ tell us that $y(0)=a_0$ and $y'(0)=a_1$ , so we get the recurrence

$\begin{cases}a_0=a_0\\\\a_1=a_1\\\\a_n=\dfrac{a_{n-1}}n&\text{for }n\ge2\end{cases}$

We can solve explicitly for $a_n$ quite easily:

$a_n=\dfrac{a_{n-1}}n\implies a_{n-1}=\dfrac{a_{n-2}}{n-1}\implies a_n=\dfrac{a_{n-2}}{n(n-1)}$

and so on. Continuing in this way we end up with

$a_n=\dfrac{a_1}{n!}$

so that the solution to the ODE is

$y(x)=\displaystyle\sum_{n\ge0}\dfrac{a_1}{n!}x^n=a_1+a_1x+\dfrac{a_1}2x^2+\cdots=a_1e^x$

We also require the solution to satisfy $y(0)=a_0$ , which we can do easily by adding and subtracting a constant as needed:

$y(x)=a_0-a_1+a_1+\displaystyle\sum_{n\ge1}\dfrac{a_1}{n!}x^n=\underbrace{a_0-a_1}_{C_2}+\underbrace{a_1}_{C_1}\displaystyle\sum_{n\ge0}\frac{x^n}{n!}$

4 0

3 years ago

Write in simplest radical form: √208

Arte-miy333 [17]

$208:2=104\\104:2=52\\52:2=26\\26:2=13\\13:13=1\\\\208=2\cdot2\cdot2\cdot2\cdot13=2^2\cdot2^2\cdot13\\\\\sqrt{208}=\sqrt{2^2\cdot2^2\cdot13}=\sqrt{2^2}\cdot\sqrt{2^2}\cdot\sqrt{13}=2\cdot2\cdot\sqrt{13}=4\sqrt{13}\\\\Answer:\ \sqrt{208}=4\sqrt{13}\\\\Used:\\\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\\sqrt{a^2}=a$

6 0

3 years ago